Question: The equation of an ellipse $E$ is $\dfrac {(y-4)^{2}}{49}+\dfrac {(x-6)^{2}}{16} = 1$. What are its center $(h, k)$ and its major and minor radius?
Explanation: The equation of an ellipse with center $(h, k)$ is $ \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1$ We can rewrite the given equation as $\dfrac{(x - 6)^2}{16} + \dfrac{(y - 4)^2}{49} = 1 $ Thus, the center $(h, k) = (6, 4)$ $49$ is bigger than $16$ so the major radius is $\sqrt{49} = 7$ and the minor radius is $\sqrt{16} = 4$.